# 28 Moments and Moment Generating Functions

## 28.1 Definitions of Moments

### 28.1.1 Definition: General Definition of Moments

The $$k^{th}$$ moment of a random variable $$X$$ about some point $$c$$ is defined to be $$E[(X-c)^k]$$.

There are two moments that are of particular use in statistics. First, the moment of $$X$$ about the origin; second, the moment of $$X$$ about the mean.

### 28.1.2 Definition: Ordinary Moments

The $$k^{th}$$ moment of a random variable $$X$$ about the origin is defined to be $$E[(X-0)^k] = E(X^k)$$.

### 28.1.3 Definition: Central Moments

The $$k^{th}$$ moment of a random variable $$X$$ about the mean $$\mu$$ is defined to be $$E[(X-\mu)^k]$$.

Using these definitions we can derive the first three central moments as follows:

\begin{aligned} E[(X-\mu)^1] &= E(X - \mu) \\ &= E(X) - \mu \\ &= E(X) - E(X) \\ \\ E[(X-\mu)^2] &= E[(X-\mu)(X-\mu)] \\ &= E(X^2-\mu X-\mu X+\mu^2) \\ &= E(X^2-2\mu X+\mu^2) \\ &= E(X^2) - E(2\mu X) + E(\mu^2) \\ &= E(X^2) - 2\mu E(X) + \mu^2 \\ &= E(X^2) - 2\mu\cdot\mu + \mu^2 \\ &= E(X^2) - 2\mu^2 + \mu^2 \\ &= E(X^2) - \mu^2 \\ &= E(X^2) - E(X)^2 \\ \\ \\ E[(X-\mu)^3] &= E[(X-\mu)(X-\mu)(X-\mu)] \\ &= E[(X^2-2\mu X+\mu^2)(X-\mu)] \\ &= E(X^3-\mu X^2-2\mu X^2+2\mu^2X+\mu^2X+\mu^3) \\ &= E(X^3-3\mu X^2+3\mu^2X-\mu^3) \\ &= E(X^3) - E(3\mu X^2) + E(3\mu^2X) - E(\mu^3) \\ &= E(X^3) - 3\mu E(X^2) + 3\mu^2E(X) - \mu^3 \\ &= E(X^3) - 3\mu E(X^2) + 3\mu^3 - \mu^3 \\ &= E(X^3) - 3\mu E(X^2) + 2\mu^3 \end{aligned}

It should be noticed that with all of these results, the moment about the mean can be evaluated by finding the ordinary moments. Thus, if we can find a consistent way to generate ordinary moments , we may use these results to find various parameters of a distribution.

## 28.2 Moment Generating Functions

### 28.2.1 Definition: Moment Generating Function

The moment generating function of a random variable, denoted $$M_X(t)$$, is defined to be:

$M_X(t) = E(e^{tX})$

The moment generating function of $$X$$ is said to exist if for any positive constant $$c,\ M_X(t)$$ is finite for $$|t|<c$$. The definition can be expanded to

\begin{aligned} M_X(t) &= E(e^{tX}) \\ &= \sum\limits_{i=1}^{\infty}e^{tx_i}p(x_i) \\ ^{[1]} &= \sum\limits_{i=1}^{\infty}[\frac{(tx_i)^0}{0!}+\frac{(tx_i)^1}{1!} + \frac{(tx_i)^2}{2!}+\frac{(tx_i)^3}{3!}+\cdots]p(x_i) \\ &= \sum\limits_{i=1}^{\infty}[1+tx_i+\frac{(tx_i)^2}{2!} + \frac{(tx_i)^3}{3!}+\cdots]p(x_i) \\ &= \sum\limits_{i=1}^{\infty}[p(x_i)+tx_ip(x_i) + \frac{(tx_i)^2}{2!}p(x_i)+\frac{(tx_i)^3}{3!}p(x_i) + \cdots] \\ &= \sum\limits_{i=1}^{\infty}p(x_i)+\sum\limits_{i=1}^{\infty}tx_ip(x_i) + \sum\limits_{i=1}^{\infty}\frac{(tx_i)^2}{2!}p(x_i) + \sum\limits_{i=1}^{\infty}\frac{(tx_i)^3}{3!}p(x_i) + \cdots \\ &= \sum\limits_{i=1}^{\infty}p(x_i)+t\sum\limits_{i=1}^{\infty}x_ip(x_i) + \frac{t^2}{2!}\sum\limits_{i=1}^{\infty}x_i^2p(x_i) + \frac{t^3}{3!}\sum\limits_{i=1}^{\infty}x_i^3p(x_i)+\cdots \\ &= 1 + tE(X) + \frac{t^2}{2!}E(X^2) + \frac{t^3}{3!} E(X^3) + \cdots \end{aligned}

1. Taylor Series Expansion: $$e^x=\frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}\cdots = 1+x+\frac{x^2}{2!}+\cdots$$

### 28.2.2 Theorem: Extraction of Moments from Moment Generating Functions

Let $$M_X^{(k)}(t)$$ denote the $$k^{th}$$ derivative of $$M_X(t)$$ with respect to $$t$$. Then $$M_X^{(k)}(0)=E(X^k)$$.

Proof:

\begin{aligned} M_X(t) &= 1 + tE(X) \\ &= \frac{t^2}{2!}E(X^2) + \frac{t^3}{3!} + \cdots \\ \\ \\ M_X^{(1)}(t) &= 0 + E(X) + \frac{2t}{2!}E(X^2) + \frac{3t^2}{3!}E(X^3) + \cdots \\ &= E(X) + tE(X^2) + \frac{t^2}{2!}E(X^3) + \cdots \\ \\ \\ M_X^{(2)}(t) &= 0 + E(X^2) + \frac{2t}{2!}E(X^3) + \frac{3t^2}{3!}E(X^4) + \cdots \\ &= E(X^2) + tE(X^3) + \frac{t^2}{2!}E(X^4) + \cdots \\ \vdots \\ \\ \\ M_X^{(k)}(t) &= 0 + E(X^k) + \frac{2t}{2!}E(X^{k+1}) + \frac{3t^2}{3!}E(X^{k+2}) + \cdots \\ &= E(X^k) + tE(X^{k+1}) + \frac{t^2}{2!}E(X^{k+2}) + \cdots \\ \\ \\ M_X^{(1)}(0) &= 0 + E(X) + \frac{2\cdot 0}{2!}E(X^2) + \frac{3\cdot 0t^2}{3!}E(X^3) + \cdots \\ &= E(X)\\ \\ \\ M_X^{(2)}(0) &= 0 + E(X^2) + \frac{2\cdot 0}{2!}E(X^3) + \frac{3\cdot 0^2}{3!}E(X^4) + \cdots \\ &= E(X^2) \\ \\ \\ \vdots \\ \\ \\ M_X^{(0)}(t) &= 0 + E(X^k) + \frac{2\cdot 0}{2!}E(X^{k+1}) + \frac{3\cdot 0^2}{3!}E(X^{k+2}) + \cdots \\ &= E(X^k) \end{aligned}